Mathematics - Inverse Trigonometric Functions Question with Solution | TestHub

Put t=${\tan }^{-1}x\in (-\pi /2,\pi /2)$, so ${\cot }^{-1}x=\pi /2-t$. The equation gives t²-$\pi t/2-3{\pi }^2/16=0$, with t=-$\pi /4$ or 3$\pi /4$. Only -$\pi /4$ is in range, giving one x=-1.